The solution of the partial differential equation $\frac{{\partial u}}{{\partial t}} = \alpha \frac{{{\partial ^2}u}}{{\partial {x^2}}}$ is of the form

Question

The solution of the partial differential equation $\frac{{\partial u}}{{\partial t}} = \alpha \frac{{{\partial ^2}u}}{{\partial {x^2}}}$ is of the form

TuteeHUB · Accepted Answer

$C{e^{kt}}{C_1}\cos \left( {\sqrt {\frac{k}{\alpha }} } \right)x + {C_2}\sin \left( { - \sqrt {\frac{k}{\alpha }} } \right)x$

TuteeHUB · Answer

$C\cos \left( {kt} \right){C_1}{e^{\left( {\sqrt {\frac{k}{\alpha }} } \right)x}} + {C_2}{e^{ - \left( {\sqrt {\frac{k}{\alpha }} } \right)x}}$

TuteeHUB · Answer

$C{e^{kt}}[{C_1}{e^{\left( {\sqrt {\frac{k}{\alpha }} } \right)x}} + {C_2}{e^{ - \left( {\sqrt {\frac{k}{\alpha }} } \right)x}}]$

TuteeHUB · Answer

$C\sin \left( {kt} \right){C_1}\cos \left( {\sqrt {\frac{k}{\alpha }} } \right)x + {C_2}\sin \left( { - \sqrt {\frac{k}{\alpha }} } \right)x$

1.	The solution of the partial differential equation \(\frac{{\partial u}}{{\partial t}} = \alpha \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is of the form
A.	\(C\cos \left( {kt} \right){C_1}{e^{\left( {\sqrt {\frac{k}{\alpha }} } \right)x}} + {C_2}{e^{ - \left( {\sqrt {\frac{k}{\alpha }} } \right)x}}\)
B.	\(C{e^{kt}}[{C_1}{e^{\left( {\sqrt {\frac{k}{\alpha }} } \right)x}} + {C_2}{e^{ - \left( {\sqrt {\frac{k}{\alpha }} } \right)x}}]\)
C.	\(C{e^{kt}}{C_1}\cos \left( {\sqrt {\frac{k}{\alpha }} } \right)x + {C_2}\sin \left( { - \sqrt {\frac{k}{\alpha }} } \right)x\)
D.	\(C\sin \left( {kt} \right){C_1}\cos \left( {\sqrt {\frac{k}{\alpha }} } \right)x + {C_2}\sin \left( { - \sqrt {\frac{k}{\alpha }} } \right)x\)
Answer» C. \(C{e^{kt}}{C_1}\cos \left( {\sqrt {\frac{k}{\alpha }} } \right)x + {C_2}\sin \left( { - \sqrt {\frac{k}{\alpha }} } \right)x\)

The solution of the partial differential equation \(\frac{{\partial u}}{{\partial t}} = \alpha \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is of the form

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